PhysicsCBSE Class 11 · Systems of Particles and Rotational Motion

Centre of Mass and its Motion Class 11

Updated 17 October 2026 · Fact-checked

The centre of mass of particles is the mass-weighted average position, R = Σmᵢrᵢ/M. Its velocity is V = Σmᵢvᵢ/M = P/M and it moves as if all the mass were concentrated there and all external forces acted there: M·A = F_ext. Internal forces cannot change its motion, so with no external force the centre of mass keeps its state of rest or uniform motion.

Understand Centre of Mass and its Motion

Position. For particles of masses $m_{i}$ at positions $r_{i}$ , the centre of mass (CM) is
$ $R = \frac{\sum m _{i} r _{i}}{M}, M = \sum m_{i},$ $
and for a continuous body $R = \frac{1}{M} \int r d m$ . In components, $X = \frac{\sum m _{i} x _{i}}{M}$ and so on. The CM of a symmetric uniform body lies at its centre of symmetry (the centre of a ring, a rod, a disc, a sphere), even where there is no mass (a ring). For a uniform triangular lamina it is at the point where the medians meet.

Removing a part. Treat a hole as a body of *negative* mass: $M \cdot 0 = (M - m) x_{r es t} + m x_{h o l e}$ , so $x_{r es t} = - \frac{m x _{h o l e}}{M - m}$ .

Variable density. For a rod with linear density $λ (x)$ , $X = \frac{\int x λ d x}{\int λ d x}$ .

Motion of the CM. Differentiating, $V = \frac{\sum m _{i} v _{i}}{M} = \frac{P}{M}$ , and
$ $M A = F_{e x t} .$ $
Internal forces occur in equal and opposite pairs and add to zero. Consequences:

If $F_{e x t} = 0$ , the CM stays at rest or moves uniformly in a straight line, whatever the internal motion (explosions, a person walking on a boat on still water, a spring oscillating).
The CM of a projectile that explodes in flight continues along the original parabola until a fragment lands.
A change $Δ R = \frac{\sum m _{i} Δ r _{i}}{M}$ : if one particle moves, the CM moves by its fraction of the mass.

CM frame. Velocities relative to the CM add up (with mass weights) to zero, and the total kinetic energy is the energy of CM motion, $\frac{1}{2} M V^{2}$ , plus the kinetic energy about the CM.

Centre of Mass and its Motion: formulas and key facts

CM of particles: $R = \frac{\sum m _{i} r _{i}}{\sum m _{i}}$; Weighted average.
CM velocity: $V = \frac{\sum m _{i} v _{i}}{M} = \frac{P}{M}$; Total momentum over mass.
CM acceleration: $M A = F_{e x t}$; Internal forces drop out.
Hole as negative mass: $x_{r es t} = - \frac{m x _{h o l e}}{M - m}$; Origin at the CM of the full body.
Continuous body: $X = \frac{\int x d m}{\int d m}$; Use $d m = λ d x$ .
Kinetic energy: $K = \frac{1}{2} M V^{2} + K_{ab o u t C M}$; Splits into two parts.

How to solve Centre of Mass and its Motion questions

Pick an origin, tabulate masses and positions, then take the mass-weighted average along each axis.

1Choose an origin and axes that make coordinates simple (often at one body or at the centre).
2List each mass with its coordinates (replace a symmetric part by a point mass at its centre).
3For a cut-out, subtract it as a negative mass.
4Compute ΣmX/M and ΣmY/M.
5For motion problems, use M·A = F_ext and ignore internal forces.
6If no external force acts, the CM does not move: shifts of the parts are related by Σm·Δx = 0.

Centre of Mass and its Motion in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Two skaters on a rink

Suppose two skaters of masses 50 kg and 70 kg hold the ends of a 6 m rope on frictionless ice and pull themselves together.

Where do they meet, measured from the 50 kg skater's start?

Try it first, then show the solutionSolution

No external horizontal force, so the CM stays fixed.
CM from the 50 kg skater: (70 × 6)/120 = 3.5 m.

Answer: 3.5 m from the 50 kg skater's starting point.

Memory hook: They meet at the CM, nearer the heavier person.

A loaded plank

Suppose a uniform plank 4 m long and 20 kg has a 10 kg box fixed at one end.

Where is the CM of plank and box from the box end?

Try it first, then show the solutionSolution

(10 × 0 + 20 × 2)/30 = 1.33 m.

Answer: 1.33 m from the box end.

Memory hook: The CM shifts towards the heavy end.

Tricks to remember Centre of Mass and its Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

Hole = negative mass
Subtract the removed piece as a negative point mass at its own centre.
Use it when: Plates with a cut-out.
No external force, no CM move
Everything inside can move, but the CM does not.
Use it when: Boat and man, explosion.
Move one, CM moves a fraction
ΔR = m Δx / M.
Use it when: Quick shifts.
Symmetry first
The CM of a uniform symmetric body is on the axis or centre of symmetry.
Use it when: Saving time.

Centre of Mass and its Motion: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: CM on a line · 2 marks · Very short answerBoard level

Masses 1 kg, 2 kg and 3 kg lie at x = 0, 3 m and 6 m. Find their centre of mass.

Try it first, then show the solutionSolution

X = (1 × 0 + 2 × 3 + 3 × 6)/(1 + 2 + 3). (1)
= 24/6 = 4 m. (1)

Answer: 4 m from the origin.

Example 2: Rod with a particle · 3 marks · Short answerBoard level

A uniform rod of mass 3 kg and length 1.2 m has a 1 kg particle attached at one end. Find the CM from the particle's end.

Try it first, then show the solutionSolution

Rod's CM is at its midpoint, 0.6 m from either end. (1)
Take the origin at the particle: X = (1 × 0 + 3 × 0.6)/4. (1)
= 0.45 m from the particle. (1)

Answer: 0.45 m from the particle's end.

Example 3: Man on a boat · 3 marks · Short answerHarder board level

A 60 kg man at one end of a 120 kg boat of length 6 m on still water walks 4 m towards the other end (relative to the boat). Ignoring water resistance, find how far the boat moves and how far the man moves relative to the water.

Try it first, then show the solutionSolution

No external horizontal force, so the CM is fixed: let the boat move d backwards; the man moves 4 − d forwards relative to the water. (1)
60(4 − d) = 120 d gives d = 240/180 = 1.33 m. (1)
Man moves 4 − 1.33 = 2.67 m relative to the water. (1)

Answer: Boat 1.33 m backwards; man 2.67 m forwards.

Example 4: Disc with a hole · 4 marks · Long answerHarder board level

A uniform disc of radius R = 0.6 m has a circular hole of radius R/2 cut out with its centre at a distance R/2 from the disc's centre. Find the CM of the remaining plate.

Try it first, then show the solutionSolution

Mass is proportional to area: the hole removes 1/4 of the mass M. (1)
Origin at the disc's centre: M × 0 = (3M/4) x + (M/4)(R/2). (1)
x = −(R/8)/(3/4) = −R/6. (1)
x = −0.6/6 = −0.1 m: 0.1 m from the centre, on the side opposite the hole. (1)

Answer: 0.1 m from the centre, away from the hole.

Example 5: Pulling a system with a spring · 4 marks · Long answerHOTS / competency

Blocks of 2 kg and 3 kg lie on a frictionless table joined by a light spring, at x = 0 and x = 0.5 m. A constant horizontal force of 10 N acts on the 2 kg block along the line joining them, away from the other block (starting from rest). Find the acceleration of the CM, the position of the CM after 3 s, the total momentum at 3 s, and say whether the spring force changes these answers.

Try it first, then show the solutionSolution

A_cm = F/M = 10/5 = 2 m/s². (1)
Take +x towards the 3 kg block. X(0) = (2 × 0 + 3 × 0.5)/5 = 0.3 m. The force acts along −x, so the CM displacement in 3 s = −½ × 2 × 9 = −9 m, giving X = 0.3 − 9 = −8.7 m. (1)
P = Ft = 10 × 3 = 30 kg m/s along −x. (1)
The spring force is internal, so these results are unchanged; it only affects the motion about the CM. (1)

Answer: 2 m/s² (along −x); X = −8.7 m (8.7 m on the far side of the origin); P = 30 kg m/s (along −x); no change.

Where students lose marks in Centre of Mass and its Motion

The slips examiners see most often, and what to write instead.

Mistake: Putting the origin carelessly and forgetting a coordinate.
Using only x for a 2-D system.
Do this: Compute X and Y separately.
Usually costs: 1 mark.
Mistake: Not treating a hole as negative mass.
Subtracting areas without mass weights.
Do this: Mass ∝ area for a uniform plate; remove it as a negative mass.
Usually costs: 2 marks.
Mistake: Assuming the CM moves when internal forces act.
Confusing internal and external forces.
Do this: Only external forces change the CM motion.
Usually costs: Concept mark.
Mistake: Using the whole rod length as the distance to the CM.
Not using the midpoint.
Do this: A uniform rod has its CM at the centre.
Usually costs: 1 mark.

Exam tips

State the origin and the axes in the first line.
For cut-outs write the equation M·0 = (M − m)x + m x_hole.
For man-boat questions write "no external horizontal force, so CM fixed".
Quote results with the side of the centre in your answer.
Give both coordinates for 2-D systems.

Practice questions on Centre of Mass and its Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQHarder board level
Particles of 1 kg at x = 0 and 3 kg at x = 4 m lie on a line. If the 1 kg particle is moved 2 m towards the other, the CM moves by
(a) 0.5 m (b) 1 m (c) 2 m (d) 0.25 m
Hint
ΔR = mΔx/M.
Answer
(a): ΔR = 1 × 2/4 = 0.5 m.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When a man walks on a floating raft, the CM of man plus raft stays at rest if there is no horizontal force from the water.
Reason (R): Forces between the man and the raft are internal, so they do not change the CM velocity.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
External force zero.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Masses of 2 kg, 3 kg and 5 kg are at (0, 0), (4, 0) and (0, 6) metres. Find the CM.
Hint
Average x and y separately.
Answer
X = (3 × 4)/10 = 1.2 m; Y = (5 × 6)/10 = 3 m; CM = (1.2 m, 3 m).
Q4 · 3 marks · Short answerHarder board level
A 2 kg particle moves at 3 m/s eastwards and a 4 kg particle at 6 m/s westwards. Find the velocity of the CM and the kinetic energy of the system about the CM.
Hint
Take east as positive.
Answer
V_cm = (6 − 24)/6 = −3 m/s (3 m/s west). Relative to the CM: 6 m/s east and 3 m/s west, so K = ½ × 2 × 36 + ½ × 4 × 9 = 54 J (total K = 81 J; CM motion 27 J).
Q5 · 4 marks · Long answerHOTS / competency
A thin rod of length 3 m has a linear density λ = 2x kg/m, where x is measured from one end. Find its mass and the position of its centre of mass. Where would you put a support to balance it?
Hint
Integrate λ dx and xλ dx.
Answer
M = ∫₀³ 2x dx = 9 kg. ∫xλ dx = ∫₀³ 2x² dx = 18. X = 18/9 = 2 m from the light end; the support goes there (at the CM).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.CM of particles?Show answer
R = Σmᵢrᵢ/M (mass-weighted average of positions).
2.What changes the CM motion?Show answer
Only external forces.
3.CM of a uniform rod?Show answer
At its centre.
4.CM of a ring?Show answer
At its centre, where there is no mass.
5.M·A equals?Show answer
Net external force.
6.CM of a triangular lamina?Show answer
Intersection of medians.

Centre of Mass and its Motion: frequently asked questions

Can the CM be outside the body?

Yes, e.g. for a ring or a hollow shape.

Why does an exploding shell's CM follow the same path?

The explosion forces are internal, so only gravity acts externally on the CM.

Is CM the same as the centre of gravity?

In a uniform gravitational field they coincide.

How do you find the CM of an irregular lamina?

Hang it from two points; the CM lies below the points of suspension where the plumb lines cross.

What is the velocity of the CM of an isolated system?

Constant; it equals the total momentum divided by the total mass.